Equation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 64 Accepted Submission(s): 20
Problem Description
Gorwin is very interested in equations. Nowadays she gets an equation like this
x1+x2+x3+?+xn=n,
and here
0≤xi≤nfor1≤i≤nxi≤xi+1≤xi+1for1≤i≤n?1
For a certain n,
Gorwin wants to know how many combinations of xi
satisfies above condition.
For the answer may be very large, you are expected output the result after it modular
m.
Input
Multi test cases. The first line of the file is an integer
T
indicates the number of test cases.
In the next T
lines, every line contain two integer n,m.
[Technical Specification]
1≤T<20
1≤n≤50000
1≤m≤1000000000
Output
For each case output should occupies one line, the output format is Case #id: ans, here id is the data number starting from 1, ans is the result you are expected to output.
See the samples for more details.
Sample Input
2 3 100 5 100
Sample Output
Case #1: 2 Case #2: 3
Source
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5185
题目大意：问按照题目所给的公式，有多少种不同的方法得到n，方法数对m取余
题目分析：因为n比较大，直接背包，时间空间都不允许，考虑公式性质，最大的情况下获得n，即1～ma求和，ma * (ma + 1) / 2 == n
化简可以得到ma = (sqrt(8n + 1) - 1) / 2，时间空间复杂度均化为nsqrt(n)，考虑dp[i][j]表示前i个数字合成数字j的种类数，则转移方程为
dp[i][j] = dp[i - 1][j - i] + dp[i][j - i]，前i个数字合成j的种类数等于合成j－i时放了i和没放i两种情况的和，dp[0][0] = 1
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
int dp[317][50001];
int main()
{
int T, n, m;
scanf("%d", &T);
for(int ca = 1; ca <= T; ca++)
{
dp[0][0] = 1;
scanf("%d %d", &n, &m);
int ans = 0, ma = (sqrt(8 * n + 1) - 1) / 2;
for(int j = 1; j <= n; j++)
for(int i = 1; i <= min(j, ma); i++)
dp[i][j] = (dp[i][j - i] + dp[i - 1][j - i]) % m;
for(int i = 1; i <= ma; i++)
ans = (ans + dp[i][n]) % m;
printf("Case #%d: %d\n", ca, ans);
}
}